 What about equilibrium? What does that mean? In the context of forces, it means that everything is balanced out. The sum of the forces in both x and y are 0. And therefore, there is no acceleration in x or in y. Let's think about this in terms of a diagram. How can you get equilibrium? One situation is I just have no forces. So if there are no forces, then the sum of the forces is 0 in both cases. But I can have forces and still have it in equilibrium. So let's look at that. Well, if I add in a single force, that can never be in equilibrium. In that particular case, there's nothing to balance it out. So you're always going to have a net force. In order to have equilibrium, there's got to be something on the other side to balance it out. And those two forces then would have to be exactly equal. Let's take a quick look at the equation that would be necessary to solve this. If this is, say, force 1, and I call this one force 2, then my sum of the forces in x is going to be equal to plus f1 minus f2. Notice we put the direction of f2 in as the negative. And just leave it as a value in there. In order for that to be in equilibrium, those two things have to be equal to 0 once I add them up. If I do the algebra, what I see is that means that f1 has to be equal to f2. I added f2 to both sides of the equation. If I was given this in terms of numbers, if f1 is equal to 30 newtons, f2 also has to be equal to 30 newtons. Notice I don't say f2 is equal to negative 30 newtons. f1 is 30 newtons to the right. f2 is 30 newtons to the left. Oftentimes, in your problems, you're given things in terms of left, right, up, down. And therefore, you have to go in and figure out where that is on the diagram left, right, and then is that a plus or a minus sign in your equation. But when you solve for the actual force, f2 is a magnitude of that force, is 30 newtons. Now, if I were going to go through and properly put all of my symbols in here, I would probably want to go through and say, plus the magnitude of f1 minus the magnitude of f2, which means the magnitude of f1 is equal to the magnitude of f2. The magnitude of f1 is 30. The magnitude of f2 is 30. And the left right tells us our directions. Most of the times, the textbooks just use a bold-faced height for the force. And so a lot of times in our problems, rather than cluttering it up with all of these magnitude signs, we just use f1 as the magnitude of that force, how much force I have. And don't forget, it's got to have its units on there. You expect units of newtons. You can use kilograms meters per seconds in there as well. Now, if I want to be complete, I'd also come through here and look at the y equation. Well, in this equation, there are no forces. And so, of course, that no forces equals up to 0. Now, here it's really easy to see. I've got two forces that exactly balance each other. Let's look at one more case, which sometimes confuses students. Let's say I have a box. I've got two separate forces pulling to the right and one force pulling to the left. Students often take a look at this and say, oh, no, that's not an equilibrium because I've got two over here and I've got one over here. Well, in order to determine if it's an equilibrium, we'd really have to know the numbers. So let me give you an example. If f1 is equal to 10 newtons and f2 is equal to 20 newtons, if f3 is equal to 30 newtons, then when we write our equation, I've got plus f1 plus f2 minus f3 because these two to the right, that one's to the left. And once I plug in my numbers, that's plus 10 newtons plus 20 newtons minus 30 newtons. That is indeed equal to 0. And therefore, this is a problem with equilibrium. However, if I had f1 equal to 10 newtons, f2 equal to 10 newtons and f3 equal to 30 newtons, well, you should be able to see here without me having to write out the entire equation. I've got a total of 20 newtons pulling to the right, but I've got 30 newtons pulling to the left. It's not an equilibrium. Specifically, if I were to work out the full equations, the sum of the forces in the x is going to be minus 10 newtons. 10 newtons towards the left is the extra force that's not balanced out by everything else. So a diagram like this could be an equilibrium, or it might not be an equilibrium. You might have more force to the right, or you might have more force to the left. It just depends on what those numbers are. If we took a situation where we're told it is an equilibrium, then that can actually help us find an unknown force. So let's just take an example here where I've got a force f1, f2, and f3, very much like my last example. Write this out as the sum of the forces in x plus f1 plus f2 minus f3. And because it told me it's an equilibrium, I know those have to balance out to be 0. Now if I'm given two of these three forces, maybe I'm told that f1 is 20 newtons. We'll use some different numbers this time. And f3 is equal to 50 newtons. I could come back and say, what's my unknown f2? Well, these are simple enough numbers. You can probably conceptually picture out. If I'm going to have this in equilibrium, I have to have the same amount over here pulling to the left as I have pulling to the right. So if I've got 50 over here and only 20 here, well, what do I need to make up to equal out to that 50? It's going to be 30. Let's take a look in our algebra. Again, if we do the simple algebra here in this case, you should be able to apply it in later problems. So in this case, we've got 20 newtons in the positive direction plus f2, which I don't know yet, minus my 50 newtons equals 0. Well, you do the math here, and that's f2. Move over to this side. Add the 50 over there. Subtract the 20. And what I see is 30 newtons. A value of 30 newtons is what's needed here in order for the system to be in equilibrium. Again, this is a pretty simple case where if you looked at the numbers, you could probably figure it out. But we can also use this in much more complicated problems where we've got all sorts of things going on here. You write out your equation and say, because it's in equilibrium, it must be equal to 0. And then you're able to solve for the unknown quantity.